A function y=f(x) is even when f(- x)=f(x) for each x in the domain of f(x).
A function y=f(x) is odd when f(- x)=- f(x) for each x in the domain of f(x).
With this in mind, we can start to verify the different statements given in the exercise.
Statement A.
A. Any constant multiple of an even function is even.
To check this, let's consider the even function f(x)=f(- x) and the function g(x)=af(x) with a as a real number. This way, g(x) is a constant multiple of an even function. Let's now check if g(x) is even to verify the statement.
As we can see, g( - x)= - g(x). Then, g(x) is odd and the statement is true.
Statement C.
C. The sum or difference of two even functions is even.
To check for this, we will use the two even functions f(x)=f(- x) and g(x)=g(- x). Furthermore, we will use h(x)=f(x) ± g(x). Let's check if h(x) is even.
Since h(x) = h(- x), h(x) is even and the statement is true.
Statement D.
D. The sum or difference of two odd functions is odd.
We will consider two odd functions f(x) and g(x), such that f(- x)=- f(x) and g(- x) = - g(x). Furthermore, we will use the function h(x)=f(x) ± g(x) so we can check if h(x) is odd.
Since h(- x)= - h(x), h(x) is odd, the statement is true.
Statement E.
E. The sum or difference of an even and an odd functions is odd.
We will consider the even function f(x)=f(- x) and the odd function g(x), such that g(- x)=- g(x). Furthermore, we will use the function h(x)=f(x) ± g(x) so we can check if h(x) is odd.
Note that - f(x) ± g(x) is not equal to h(x)=f(x) ± g(x). Therefore, h(x) is not odd, as h(- x) ≠ - h(x). Therefore, the statement is false. In fact, the sum or difference of an even and an odd function is neither even nor odd.
